$2^n + 1$ is prime $ implies n$ is a power of $2$












2














I was wondering why I can't do this this way by proof by contradiction of the contrapositive. So I want to prove

$$2^n + 1 quad text{is prime} implies n = 2^k quad text{for some} kin mathbb{Z}.$$

By contrapositive,

$$n neq 2^k quad text{for all} k in mathbb{Z} implies 2^n + 1 quad text{is composite.}$$



Then I want to prove by contradiction:

Suppose $n neq 2^k quad forall kin mathbb{Z}$ and $2^n + 1$ is prime.

Why can't I give one counterexample to prove that this is false (contradiction)?

Since if $n:= 3$ then $2^3 + 1 = 9$ which is composite, hence, $2^n + 1$ being prime (hypothesis) cannot be true, so by proof by contradiction, $2^n +1$ must be composite?






elementary-number-theory proof-verification prime-numbers fermat-numbers







share|cite|improve this question




















  • 2




    Because you have to prove it for all $k in mathbb{Z}$.
    – Math Lover
    Aug 15 '17 at 4:18












  • Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
    – hardmath
    Aug 15 '17 at 4:33






  • 1




    You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
    – Jyrki Lahtonen
    Aug 15 '17 at 4:45












  • @Math Lover I think it should be for all $kinmathbb N$.
    – Michael Rozenberg
    Aug 15 '17 at 4:53






  • 3




    Possible duplicate of Fermat primes relation to $2^n+1$
    – Gerry Myerson
    Aug 19 '17 at 22:26
















2














I was wondering why I can't do this this way by proof by contradiction of the contrapositive. So I want to prove

$$2^n + 1 quad text{is prime} implies n = 2^k quad text{for some} kin mathbb{Z}.$$

By contrapositive,

$$n neq 2^k quad text{for all} k in mathbb{Z} implies 2^n + 1 quad text{is composite.}$$



Then I want to prove by contradiction:

Suppose $n neq 2^k quad forall kin mathbb{Z}$ and $2^n + 1$ is prime.

Why can't I give one counterexample to prove that this is false (contradiction)?

Since if $n:= 3$ then $2^3 + 1 = 9$ which is composite, hence, $2^n + 1$ being prime (hypothesis) cannot be true, so by proof by contradiction, $2^n +1$ must be composite?






elementary-number-theory proof-verification prime-numbers fermat-numbers







share|cite|improve this question




















  • 2




    Because you have to prove it for all $k in mathbb{Z}$.
    – Math Lover
    Aug 15 '17 at 4:18












  • Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
    – hardmath
    Aug 15 '17 at 4:33






  • 1




    You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
    – Jyrki Lahtonen
    Aug 15 '17 at 4:45












  • @Math Lover I think it should be for all $kinmathbb N$.
    – Michael Rozenberg
    Aug 15 '17 at 4:53






  • 3




    Possible duplicate of Fermat primes relation to $2^n+1$
    – Gerry Myerson
    Aug 19 '17 at 22:26














2












2








2







I was wondering why I can't do this this way by proof by contradiction of the contrapositive. So I want to prove

$$2^n + 1 quad text{is prime} implies n = 2^k quad text{for some} kin mathbb{Z}.$$

By contrapositive,

$$n neq 2^k quad text{for all} k in mathbb{Z} implies 2^n + 1 quad text{is composite.}$$



Then I want to prove by contradiction:

Suppose $n neq 2^k quad forall kin mathbb{Z}$ and $2^n + 1$ is prime.

Why can't I give one counterexample to prove that this is false (contradiction)?

Since if $n:= 3$ then $2^3 + 1 = 9$ which is composite, hence, $2^n + 1$ being prime (hypothesis) cannot be true, so by proof by contradiction, $2^n +1$ must be composite?






elementary-number-theory proof-verification prime-numbers fermat-numbers







share|cite|improve this question















I was wondering why I can't do this this way by proof by contradiction of the contrapositive. So I want to prove

$$2^n + 1 quad text{is prime} implies n = 2^k quad text{for some} kin mathbb{Z}.$$

By contrapositive,

$$n neq 2^k quad text{for all} k in mathbb{Z} implies 2^n + 1 quad text{is composite.}$$



Then I want to prove by contradiction:

Suppose $n neq 2^k quad forall kin mathbb{Z}$ and $2^n + 1$ is prime.

Why can't I give one counterexample to prove that this is false (contradiction)?

Since if $n:= 3$ then $2^3 + 1 = 9$ which is composite, hence, $2^n + 1$ being prime (hypothesis) cannot be true, so by proof by contradiction, $2^n +1$ must be composite?






elementary-number-theory proof-verification prime-numbers fermat-numbers




elementary-number-theory proof-verification prime-numbers fermat-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 21:52









greedoid

38.6k114797




38.6k114797










asked Aug 15 '17 at 4:15









OneGapLaterOneGapLater

297314




297314








  • 2




    Because you have to prove it for all $k in mathbb{Z}$.
    – Math Lover
    Aug 15 '17 at 4:18












  • Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
    – hardmath
    Aug 15 '17 at 4:33






  • 1




    You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
    – Jyrki Lahtonen
    Aug 15 '17 at 4:45












  • @Math Lover I think it should be for all $kinmathbb N$.
    – Michael Rozenberg
    Aug 15 '17 at 4:53






  • 3




    Possible duplicate of Fermat primes relation to $2^n+1$
    – Gerry Myerson
    Aug 19 '17 at 22:26














  • 2




    Because you have to prove it for all $k in mathbb{Z}$.
    – Math Lover
    Aug 15 '17 at 4:18












  • Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
    – hardmath
    Aug 15 '17 at 4:33






  • 1




    You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
    – Jyrki Lahtonen
    Aug 15 '17 at 4:45












  • @Math Lover I think it should be for all $kinmathbb N$.
    – Michael Rozenberg
    Aug 15 '17 at 4:53






  • 3




    Possible duplicate of Fermat primes relation to $2^n+1$
    – Gerry Myerson
    Aug 19 '17 at 22:26








2




2




Because you have to prove it for all $k in mathbb{Z}$.
– Math Lover
Aug 15 '17 at 4:18






Because you have to prove it for all $k in mathbb{Z}$.
– Math Lover
Aug 15 '17 at 4:18














Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
– hardmath
Aug 15 '17 at 4:33




Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
– hardmath
Aug 15 '17 at 4:33




1




1




You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
– Jyrki Lahtonen
Aug 15 '17 at 4:45






You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
– Jyrki Lahtonen
Aug 15 '17 at 4:45














@Math Lover I think it should be for all $kinmathbb N$.
– Michael Rozenberg
Aug 15 '17 at 4:53




@Math Lover I think it should be for all $kinmathbb N$.
– Michael Rozenberg
Aug 15 '17 at 4:53




3




3




Possible duplicate of Fermat primes relation to $2^n+1$
– Gerry Myerson
Aug 19 '17 at 22:26




Possible duplicate of Fermat primes relation to $2^n+1$
– Gerry Myerson
Aug 19 '17 at 22:26










2 Answers
2






active

oldest

votes


















4














The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.






share|cite|improve this answer





















  • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
    – OneGapLater
    Aug 15 '17 at 4:20












  • Yes, that is indeed the typical route to go about proving this statement.
    – Ziryerx
    Aug 15 '17 at 5:47



















3














Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
$$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



Clearly both factors are > 1 and thus a contradction.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2393990%2f2n-1-is-prime-implies-n-is-a-power-of-2%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.






    share|cite|improve this answer





















    • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
      – OneGapLater
      Aug 15 '17 at 4:20












    • Yes, that is indeed the typical route to go about proving this statement.
      – Ziryerx
      Aug 15 '17 at 5:47
















    4














    The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.






    share|cite|improve this answer





















    • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
      – OneGapLater
      Aug 15 '17 at 4:20












    • Yes, that is indeed the typical route to go about proving this statement.
      – Ziryerx
      Aug 15 '17 at 5:47














    4












    4








    4






    The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.






    share|cite|improve this answer












    The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 15 '17 at 4:19









    Robert IsraelRobert Israel

    319k23208458




    319k23208458












    • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
      – OneGapLater
      Aug 15 '17 at 4:20












    • Yes, that is indeed the typical route to go about proving this statement.
      – Ziryerx
      Aug 15 '17 at 5:47


















    • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
      – OneGapLater
      Aug 15 '17 at 4:20












    • Yes, that is indeed the typical route to go about proving this statement.
      – Ziryerx
      Aug 15 '17 at 5:47
















    I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
    – OneGapLater
    Aug 15 '17 at 4:20






    I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
    – OneGapLater
    Aug 15 '17 at 4:20














    Yes, that is indeed the typical route to go about proving this statement.
    – Ziryerx
    Aug 15 '17 at 5:47




    Yes, that is indeed the typical route to go about proving this statement.
    – Ziryerx
    Aug 15 '17 at 5:47











    3














    Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
    $$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



    Clearly both factors are > 1 and thus a contradction.






    share|cite|improve this answer


























      3














      Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
      $$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



      Clearly both factors are > 1 and thus a contradction.






      share|cite|improve this answer
























        3












        3








        3






        Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
        $$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



        Clearly both factors are > 1 and thus a contradction.






        share|cite|improve this answer












        Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
        $$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



        Clearly both factors are > 1 and thus a contradction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 15 '17 at 6:07









        greedoidgreedoid

        38.6k114797




        38.6k114797






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2393990%2f2n-1-is-prime-implies-n-is-a-power-of-2%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            An IMO inspired problem

            Image
            14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
            Read more

            Management

            Image
            Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
            Read more

            Has there ever been an instance of an active nuclear power plant within or near a war zone?

            Image
            36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
            Read more